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[NOTES/EM-09004]-Self Inductance and Mutual Inductance

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We define the mutual inductance for two loops and self inductance for a loop. We obtain an expression for energy of a circuit with self inductance \(L\)

1. Mutual inductance

Consider a loop carrying a current \(I_1\). The magnetic field due to the current \(I\) a loop \(\Gamma_1\) is \begin{align*} \vec{B}_1 = & \frac{\mu_1}{4\pi} \int_{\Gamma_1} I_1\frac{\overrightarrow{dl_1}\times\vec{R}}{R^3} \end{align*} If a second loop \(\Gamma_2\) is placed near to the first loop, the flux, \(\Phi_2\), of magnetic field linked with the second loop will be \begin{align*} \Phi_2 = & \int_{s_2} \bar{B}_1\cdot d\bar{S} \end{align*} The flux \(\Phi_2\) is proportional to the current in the first loop and we write \begin{equation} \Phi_2= I_1~M_{21} \end{equation} The quantity \(M_{21}\) is called mutual inductance of the two loops. Induced e.m.f. is thus obtained from \begin{align*} \mathcal E =& -\dd[\Phi_2]{t} = - M_{21}\frac{dI_1}{dt} \end{align*} It is possible to show that the mutual inductance. depends only on geometrical arrangement of the loops.

2. Self inductance

The flux \(\Phi\) of the magnetic field due to current \(I\), in a closed circuit, is also linked with the circuit. This flux will be proportional to the current \(I\). The constant of proportionality \(L\) is called self inductance of the circuit. Thus we have \(\Phi= L I\) and the induced e.m.f. will be given by \begin{align*} \mathcal E = & -L \frac{dI}{dt} \end{align*} The SI unit of self and mutual inductance is known as henery: \begin{align*} [L] = & \frac{\text{Volt}\times \text{sec}}{\text{amp}} = \text{henry} \end{align*}

3.Energy stored in a current carrying coil

The induced current opposes the cause of induction. The amount of work done per sec, \(W\), when a current \(I\) flows in a coil having self inductance \(L\), is \begin{equation} \dd{W}{t} = - \mathcal E I = L I \dd[I]{t}. \end{equation} The total work done to build a current from 0 to a value \(I\) in the is given by \begin{equation} W = \int_0^I L I \dd[I]{t} = \frac{1}{2}L I^2. \end{equation} The final answer does not depend on how slow or fast the current is built up.

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